23
$\begingroup$

Let $\varphi_t : M \rightarrow M$ be a smooth flow on a smooth manifold $M$. We may assume (although I'm not sure if this is important) that the flow preserves a smooth volume form on $M$. Given a continuous function $f : M \rightarrow \mathbb{R}$, is it true that if ALL the (partial) time averages $$ A_T f(x) := {1 \over T} \int_0^T f(\varphi_t (x)) \, dt \hskip 1cm (T > 0) $$ are smooth functions, then $f$ is also smooth?

Remark. This is obviously true for the trivial action and it is non-trivially true for the case where $M = \mathbb{R}$ and the flow is the usual flow by translations. That's all I know for now.

Motivation. In the first volume of Dunford and Schwartz, where they motivate the Ergodic theorem (page 657), they state:

What is significant and measurable in the laboratory is not the quantity $f(\phi_t(x))$ but its average value $$ {1 \over T} \int_0^T f(\varphi_t (x)) \, dt $$ computed over a certain time interval $0 \leq t \leq T$.

I'm asking whether knowing the regularity of all those average values says something about the regularity of the function, or "observable" $f$. It would be interesting to consider $f$ to be just locally integrable and then the question ties in a (very) little with Wiener's differentiation theorem.

$\endgroup$
7
  • $\begingroup$ sorry for posting it as an answer, but i don't have the right to comment. since you are not requiring anything for the limit $T \rightarrow \infty $, the normalization of the time integral by $T^{-1} $ does not really serve a purpose, does it? $\endgroup$
    – jesus
    Jul 2 '15 at 13:14
  • $\begingroup$ Not really. It only serves in pointing out that as T goes to zero, $A_T f(x) = f(x)$. The (very loose) analogy here is not with Birkhoff's ergodic theorem, but with Wiener's differentiation theorem. $\endgroup$ Jul 3 '15 at 8:05
  • $\begingroup$ I don't expect this to be true as averaging tends to improve the regularity of a function. Do you have a reference for the non-trivial result in your remark? $\endgroup$
    – K Hughes
    Jul 3 '15 at 11:01
  • 2
    $\begingroup$ The non-trivial result is equivalent to the statement that if $f$ is a continuous, real-valued function on the reals and $f(x+ T) - f(x)$ is smooth for every fixed $T$, then $f$ is smooth. I had asked this to my colleague, Jean-François Burnol and he gave a beautiful proof which, as he described, is an ode to Baire's theorem. He even proved it in the case $f$ is just assumed to be a distribution. I don't know if he published it or plans to publish it, but it is nicely written and you can ask him for a copy if you like. $\endgroup$ Jul 3 '15 at 12:55
  • 1
    $\begingroup$ I would be grateful to see a copy if possible. Thanks! $\endgroup$
    – K Hughes
    Jul 10 '15 at 13:39
3
$\begingroup$

$\textbf{Update with some corrections}$:

Unfortunately, my original answer had a serious gap. Sorry for the mix-up. The argument I gave only works when we have estimates on $A_T f(x)$ that are uniform in $T$. However, all the hard work was not useless as it made clear the properties that a counterexample would have. Without further ado, here is an example for which $A_T f(x)$ is smooth while $f(x)$ is not.

Let $M= \mathbb{R^2} =(t,x)$ with the flow $\phi_s (t, x) := (t+s, x)$.

Let $\Psi$ be the following function

$$\Psi(x) = \begin{cases} 0 & x\leq 0 \\ e^{\frac{-1}{x^2}} & x > 0 \\ \end{cases}$$

In particular, we want $\Psi$ to be a smooth function which vanishes to all orders at $0$. Let $f$ be the following function:

$$f(t,x)= \begin{cases} e^{\frac{-t^2}{(\Psi(x))^2}} & x >0 \\ 0 & x \leq 0 \end{cases} $$

The function $f$ is not smooth. In fact, it is not even continuous at $(0,0)$. To show that $A_T f(x)$ is smooth for any $T$, we must only show that it depends smoothly on $x$ near $(0,0)$. For fixed $T$>0, we have the following:

$$A_T(f(t_1,x))= \frac{1}{T} \int_{t_1}^{t_1+T} f(t,x) dt < \frac{\sqrt{ \pi}}{T} \Psi(x) = o(x^k) ~ \forall k >0 $$

From this, it follows that $\frac{\partial^k A_T f}{\partial x^k}(t,0) \equiv 0$, so the derivative of $A_T f$ exists to all orders at $(0,0)$ for $T>0$. The smoothness of $A_T f(x)$ elsewhere is immediate because $f$ is smooth away from the origin.

For an example where $f(x)$ is continuous but not smooth, we can consider the function $\hat f(t,x) = x \cdot f(t,x)$. Here, the same analysis holds except that $f$ is continuous but not differentiable at $(0,0)$.

$\textbf{Original answer with the some corrections}$

This is not a full answer, but I believe that your colleague's result does all the heavy lifting away from the fixed points of the flow, at least once you have good estimates on $A_T f(x)$. The case near a fixed point of a flow seems more difficult. I'm not sure how that works exactly, but if one can understand it in the $1$-dimensional case (i.e. when the flow has a limit point), this might give some good insight. I broke the answer into several sections to help make it more readable.

$\textbf{Finding the right coordinates}$

In this case, the smooth flow locally foliates your manifold, so around any point $p$ one can use the inverse function theorem to choose local coordinates $\{ x_0, x_1, \ldots, x_{n-1} \}$ for which the flow acts as $\phi_t(x_0) = x_0+t$ and $\phi_t(x_i) = x_i$ otherwise. By Professor Burnol's argument (which I'm black-boxing for this argument) $f( x_0, x_1, \ldots, x_{n-1})$ depends smoothly on $x_0$ when all the other coordinates are fixed.

What remains to show is that it depends smoothly on $x_1, \ldots, x_n$. We start with the case where $x_0$ is some fixed value, say 0. In this case, the $A_T f(x)$ is exactly the following:

$$ \frac{1}{T}\int_0^T f(t, x_1, \ldots, x_n) ~dt$$

For the rest of this, I will use the parameter $t$ for $x_0$ and refer to the rest of the coordinates collectively as $\mathbf{x}$.

We know that $A_T f(x)$ depends smoothly on $\mathbf{x}$ and further that $f$ depends smoothly on $t$. What remains to show is that $f$ depends smoothly on $\mathbf{x}$. We can't take derivatives with respect to $\mathbf{x}$, so we need to estimate some difference quotients. I will be using the Euclidean norm on the $\mathbf{x}$ coordinates for my estimates.

$\textbf{The uniform Lipschitz estimate}$

Take two points $\mathbf{x}^1$ and $\mathbf{x}^2$. Since $A_T f(x)$ is smooth, it is locally Lipschitz, and so we have the following estimate:

$$ \frac{1}{T \| \mathbf{x}^1- \mathbf{x}^2 \| } | \int_0^T f(t, \mathbf{x}^1) - f(t, \mathbf{x}^2) ~dt | < \mathcal{C} $$

In order for the rest of the argument to work, we must assume that this estimate is uniform in the choice of $T$, $\mathbf{x}^1$ and $\mathbf{x}^2$. This allows us to pick $T>0$ small enough so that that for our fixed $\mathbf{x}^1$ and $\mathbf{x}^2$, we have the following:

$$ \| f(t, \mathbf{x}^i) -f(s, \mathbf{x}^i) \|_{C^2} < \epsilon$$ for $0<t,s < T$.

The choice of $T$ may depend on $\mathbf{x}^1$ and $\mathbf{x}^2$. In particular, we don't expect to be able to do this uniformly in $\mathbf{x}$, but that's okay because we are only doing it for two points.

Then, we have the following estimate.

\begin{eqnarray} | \frac{1}{T} \int_0^T \frac{ f(t, \mathbf{x}^1) - f(t, \mathbf{x}^2)}{\| \mathbf{x}^1- \mathbf{x}^2 \| } ~dt | & \geq & | \frac{1}{T} \frac{ T(f(0,\mathbf{x}^1) - f(0,\mathbf{x}^2))}{\| \mathbf{x}^1- \mathbf{x}^2 \|} | \\ & & - \frac{1}{T} \left( \frac{\partial f}{\partial t} (0, \mathbf{x}^1)+ \frac{\partial f}{\partial t} (0, \mathbf{x}^2) + 2\epsilon \right) \frac{T^2}{2} \\ & \geq & | \frac{f(0,\mathbf{x}^1) - f(0,\mathbf{x}^2)}{\| \mathbf{x}^1- \mathbf{x}^2 \|}| - \left( \frac{\partial f}{\partial t} (0, \mathbf{x}^1)+ \frac{\partial f}{\partial t} (0, \mathbf{x}^2) + 2\epsilon \right) T \\ \end{eqnarray}

Letting $T$ be go to zero using the uniformity of the earlier estimate while still fixing $\mathbf{x}^1$ and $\mathbf{x}^2$ for now, our work shows the following.

$$ | \frac{f(0,\mathbf{x}^1) - f(0,\mathbf{x}^2)}{\| \mathbf{x}^1- \mathbf{x}^2 \|}| \leq \mathcal{C} $$

Notice that this $\mathcal{C}$ is the same constant as before in the estimate of $A_T f(x)$. Repeating this for arbitrary pairs of $\mathbf{x}$, this gives a uniform Lipschitz estimate on $f$ (at $t=0)$ in terms of the $\mathbf{x}$ coordinates.

$\textbf{A sketch of how to use induction for higher order regularity}$

In fact, we can repeat essentially the same argument but with three points $\mathbf{x}^1$, $\mathbf{x}^2$ and $ \lambda \mathbf{x}^1 + (1- \lambda) \mathbf{x}^2$ where instead of using the difference quotient, we use the second order difference quotient. This should yield a uniform $C^{1,1}$ estimate on $f$ in terms of the smoothness of $A_T f(x)$. With a uniform $C^{1,1}$ estimate, this shows that the function $f$ was actually differentiable in terms of $\mathbf{x}$ all along. You can induct on all orders to get smoothness of $f$ in $\mathbf{x}$. This shows that $f$ is smooth in both $t$ and $\mathbf{x}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.